(a) Pressure decreases as one ascends the atmosphere . If the density of air is `rho` ,what is the change in pressure dp over differential height dh ?
(b) Considering the pressure P to be proportional to the density find the pressure P at a height h if the pressure on the surface of the earth is `P_(o)`.
(c ) If `P_(o)=1.03xx10^(5)N//m^(-2),rho_(o)=1.29kg//m^(3)andg=9.8m//s^(2)` what height will the pressure frop to `(1)/(10)` the value at the surface of earth ?
(d) This model of the atmosphere works for relatively small distance .Identify the underlying assumption that limits the model.

Explanation

Since the air is less dense at higher up and hence pressure is also less.
(a) Consider a horizontal portion of air with cross section A and height dh.

Let the pressure on the top surface and bottom surface be P and `P_(1)+dP`. If the portion is in equilibrium then the net upward force must be balanced by the weight.
`(P+dP)A-PA=-mg" " (because"mass"="volume"xx"density")`
`therefore(dP)A=-rho(Adh)g`
`(rho=` density of air)
`thereforedp=-pgdh` ...(1)
Negative sign indicates that pressure decreases at height increases .
(b) Let the density of air on the earth.s surface be `rho_(o)`, then
Pressure `prop` density
`thereforePproprhoandP_(o)proprho_(o)`
`therefore(P)/(P_(o))=(rho)/(rho_(o))`
`thereforerho=((P)/(P_(o)))rho_(o)` ....(2)
Putting value of equation (2) in (1),
`thereforedp=-((P)/(10))rho_(o)dgh`
`therefore(dP)/(P)=-(rho_(o)g)/(P_(o))dh`

`thereforeh=(P_(o))/(rho_(o)g)ln(10)`
`thereforeh=(P_(o))/(rho_(o)g)[2.363log_(10)^(10)]`
`=((1.093xx10^(5)))/(1.22xx9.8)xx2.363(1)=0.16xx10^(5)m`
`thereforeh=16xx10^(3)m`
(d) Pressure `Pproprho` (For isothermal prosess)temperature remains constant only near the surface of earth at greater height relation `Pproprho` does not obay.Hence ,for small distances near earth `Pproprho` is obay, this is not for greater distances . This is the limitation of this model .